# Homework Help: 2D Electrical force problem

1. Jan 9, 2009

### Elbobo

1. The problem statement, all variables and given/known data
Three identical point charges, each of mass 80 g and charge +q, hang from three strings,
as in the figure.
The acceleration of gravity is 9.8 m/s2,
and the Coulomb constant is 8.98755 × 109 N · m2/C2.

If the lengths of the left and right strings
are each 11.1 cm, and each forms an angle of
47◦ with the vertical, determine the value of
q. Answer in units of μC.

2. Relevant equations

W = mg
F = (kq2) / r2
SOHCAHTOA

3. The attempt at a solution

It seemed to be a relatively easy problem, but I just keep getting it wrroooong.

So in the x-direction..

mg tan (theta) = [ (kq2) / L2 sin2 (theta) ] + [ (kq2) / (4L2 sin2) (theta)) ]
sqrt [ (mg * tan (theta) * 4L2 sin2) (theta)) / (5k) ] = q
q = 7.02271226 x 10-13 C

q * (10-6) = 7.02271226 x 10-13 micro C

Which is wrong.

EDIT: OH GOD!!!! I multiplied times 10^-6 instead of 10^6. Can someone at least check if my process is right though?

FOR CLARIFICATION:

x-component of weight on left mass: mg tan theta
length of string: L
x-length from vertical to left mass: L2 sin2 (theta)
x-length from right mass to left mass: 2*L2 sin2 (theta)

Last edited: Jan 9, 2009
2. Jan 9, 2009

### Delphi51

Elbobo, I can't understand the problem and I can't follow your work. I'm old and retired and don't want to do any arithmetic but I would love to help you with the theory.

I am guessing that you have a central charge hanging straight down. Right?
And a hanging charge on the left that is deflected 47 degrees to the left. And one on the right deflected to the right. If so, you can get a grip on the problem by looking at the forces on either the left or the right charges. Make a force diagram showing all forces acting on the charge. Any forces acting at an angle need to be separated into horizontal and vertical parts using trigonometry. Use F = ma to deduce what the total force on the charge is. Write that down separately for the horizontal forces and the vertical forces. Try to calculate each of the forces in both. You will be unable to find some of them - for example the electric force Fe because you don't know the charge it depends on. You should be able to find enough of the forces to solve one of the two force equations and thus find something new. Anything you find will be most helpful in finishing the problem!

3. Jan 9, 2009

### Elbobo

Well if it helps, all 3 charges sit on the same horizontal line.

Can't I just set the x-component of the left mass's weight equal to the x-component of the electrical force caused by the middle charge plus the x-component of the eletrical force caused by the right charge?

I assumed those two charges would combine their forces and cause a balance of weight with the left charge...

4. Jan 9, 2009

### Delphi51

Good start! But aren't both of those forces to toward the left? They would cause the left charge to accelerate to the left according to F = ma. The charge does not accelerate, so you have left something out. What holds it in place, balancing the Fe?

5. Jan 9, 2009

### Elbobo

ma = m(0) = Wx - Fe1x - Fe2x

so Wx = Fe1x + Fe2x

Isn't that what I did??

6. Jan 10, 2009

### Delphi51

I'm guessing your Wx is the weight. If so, it is a vertical force - gravity pulls straight down. The electric forces Fe1 and Fe2 are horizontal forces. You must do the horizontal forces separately from the vertical ones. Use two headings "horizontal" and "vertical" and work on them separately.

The missing force is the pull of the string, usually called "tension" or Ft. It acts in the direction of the string so it is partly horizontal and partly vertical. You know the angle so you can use trigonometry to work out the horizontal and vertical components.

7. Jan 10, 2009

### Elbobo

That's what I thought about the weight, but I couldn't think of another force along with the electrical forces. I completely forgot about tension.

Thanks a lot.

EDIT:

Oh wait a second! My first equation was still correct. Using what you gave me, I ended up with

mg tan theta = Fe1 + Fe2, which is basically what I put.

But at least I understand that my reasoning was off. I guess its being wrong was due to my multiplying times 10^-6 instead of 10^6.

Still, thanks.

8. Jan 10, 2009

### Delphi51

Most welcome!

Answers don't count for much. If you can't show clearly and convincingly how you got the whole solution, no one will want to pay you to solve their problems.